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Let $C,D$ be categories, where $C$ is small. If we have morphism $\eta$ (i.e. natural transformation) in the category of functors $D^C$, then clearly $\eta$ is a monomorphism if all components $\eta_x$, $x \in C$ are monomorphisms. The converse is true if $D$ has pullbacks:

It is well-known that $D^C$ also has pullbacks, computed pointwise. Let $\eta : F \to G$ and consider the diagonal $\Delta : F \to F \times_G F$. Then: $\eta$ is a monomorphism iff $\Delta$ is an isomorphism iff all $\Delta_x : F(x) \to F(x) \times_{G(x)} F(x)$, $x \in C$ are isomorphisms iff all $\eta_x : F(x) \to G(x)$ are monomorphisms.

Question. Can you give an example where the converse fails, i.e. a monomorphism in the category of functors which is not pointwise a monomorphism?

This is a related question.

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Related MO questions: 1, 2. – Zhen Lin Dec 25 '11 at 12:05
up vote 8 down vote accepted

Take $C = \underline{2} = 0\to1$ and let $D$ be the category generated by the following graph $$ \matrix{ & & \bullet & \mathop{\longrightarrow}\limits^a & \bullet \cr & & \scriptsize{f} \Big\downarrow {\ \ } & & {\ \ } \Big\downarrow \scriptsize{g} \cr \bullet & \mathop{\rightrightarrows}\limits^u_v & \bullet & \mathop{\longrightarrow}\limits_b & \bullet \cr } $$ with the relations $b\circ f = g\circ a$ and $b\circ u = b\circ v$.

Then $D^C$ is just the category of arrows on $D$. The map $(a,b): f\rightarrow g$ is monic, because there is no nontrival parallel pair of maps to $f$. But $b$ is not monic.

Exercise 1.11.12 on p.37 in "Borceux Handbook on Categorical Algebra I" also has a solution for a suitable $D$ in the diagram on top of that same page.

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Ah, this is very instructive, thanks! – Martin Brandenburg Jan 5 '12 at 13:36

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